Value of a such that the system has no solution/infinitely many solutions

Question

I have this system of equations here:

$Ax=b$, where:

$A=$$ \left( \begin{array}{ccc} \ 1 & 2 & 3 \\ \ 1 & 3 & 4\\ \ 1 & a & 5\\ \end{array} \right) $

$b=$$ \left( \begin{array}{c} \ a\\ \ 3\\ \ 3\\ \end{array} \right) $

I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.

I wrote my augmented matrix as:

$ \left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 1 & 3 & 4 & 3\\ \ 1 & a & 5 & 3\\ \end{array} \right) $

I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:

$ \left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 0 & 1 & 1 & 3-a\\ \ 0 & a-2 & 2 & 3-a\\ \end{array} \right) $

I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!

Answer 1

Continue reducing!

$$\left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 0 & 1 & 1 & 3-a\\ \ 0 & a-2 & 2 & 3-a\\ \end{array} \right)\longrightarrow\left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 0 & 1 & 1 & 3-a\\ \ 0 & 0 & 4-a & (a-3)^2\\ \end{array} \right)$$

Can you prove that for $\;a=4\;$ you get an incongruent system (no solution)? What happens if $\;a\neq4\;$ ?

Answer 2

After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.

Thanks to Don Antonio for fix.